Quantitative Aptitude - MCQ Practice Questions

This question asks us to find the smallest positive number that is divisible by both 12 and 18.

Step 1: Find Prime Factorization of Both Numbers

Break down each number into its prime factors.

Step 2: Identify Highest Powers of All Prime Factors

The LCM uses the highest power of each prime factor that appears.

Step 3: Calculate the LCM

Multiply the highest powers together.

The LCM of 12 and 18 is 36, which is the smallest number divisible by both numbers.

Numbers of form 7k+3: when k=5, number=7(5)+3=38.

Check: 38÷7=5 remainder 3 ✓.

Check others: 24÷7=3 rem 3 (close but let's verify 38 first), 38÷7 gives remainder 3 ✓

Let smaller odd number = x.

Then x+(x+2)=56.

So 2x+2=56, 2x=54, x=27.

Check: 27+29=56 ✓

Let numbers be 5x and 7x.

Then 5x + 7x = 120, so 12x = 120, x = 10.

Larger number = 7 × 10 = 70

Odd numbers between 10 and 30: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.

Count = 10

Even numbers from 2 to 50: 2, 4, 6, ..., 50.

This is an AP with first term = 2, last term = 50, common difference = 2.

Number of terms = 25.

Sum = 25(2+50)/2 = 25 × 26 = 650

Let original number = x.

Then x + 25% of x = 500.

So x + 0.25x = 500, 1.25x = 500, x = 400

Let smaller number = x, larger = 2x.

Then 2x - x = 45, so x = 45.

Numbers are 45 and 90.

Let integers be n-1, n, n+1.

Sum = 3n = 81, so n = 27.

The three integers are 26, 27, 28.

Largest = 28

Sum of squares = 1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55.

Formula: n(n+1)(2n+1)/6 = 5×6×$\frac{11}{6}$ = 55

LCM(9, 11) = 99 (since 9 and 11 are coprime).

Smallest three-digit multiple of 99 is 99 × 2 = 198.

20 = 2² × 5.

Sum of divisors = (1 + 2 + 4)(1 + 5) = 7 × 6 = 42.

Divisors are: 1, 2, 4, 5, 10, 20.

32 ÷ 7 = 4 remainder 4 (No). 33 ÷ 7 = 4 remainder 5 (No). 34 ÷ 7 = 4 remainder 6 (No). 32 = 7×4 + 5 = 28 + 5 (Yes). 32 ÷ 7 gives remainder 5.

The digital root is obtained by repeatedly summing the digits of a number until a single digit remains. Alternatively, the digital root of a positive integer equals the remainder when divided by 9 (or 9 if the remainder is 0).

Step 1: Sum the digits of 9876

Step 2: Sum the digits of 30

Since we've reached a single digit, the process stops.

**Answer: The digital root of 9876 is $3$ (Option D)**

Using HCF × LCM = Product of two numbers: 15 × 180 = 45 × x.

Therefore, 2700 = 45x, so x = 60.

Perfect squares between 100 and 200: 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225 (exceeds 200).

So there are 4 perfect squares.

Wait: 10² = 100 (not between), so squares are 121, 144, 169, 196.

Count = 4.

Actually checking: we need squares from 11² to 14², which is 4 numbers.

To find the remainder when $2^{50}$ is divided by 5, we use the Fermat's Little Theorem approach by finding the pattern of powers of 2 modulo 5.

Step 1: Find the pattern of powers of 2 (mod 5)

Calculate successive powers of 2 and their remainders when divided by 5:

The pattern repeats every 4 powers.

**Step 2: Apply the cycle to find $2^{50}(\mathrm{mod}5)$

Since the remainder repeats with period 4, divide the exponent by 4:

This means $2^{50}$ has the same remainder as $2^2$.

Step 3: Calculate the remainder

**Answer: The remainder is $4$ (Option D)**

We need LCM(12, 15, 20). 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.

Since 9 and 11 are coprime (HCF = 1), the number must be divisible by 9 × 11 = 99.

Let numbers be x, x+2, x+4, x+6, x+8. Sum = 5x + 20 = 120, so x = 20. Largest = 20 + 8 = 28.